Geometric distribution

Expected value and variation

Understanding how probability connects to real-world averages starts with the expected value. The definitions below describe three related measures you’ll use throughout this section.

Practice problem

Example:

A student randomly draws a bill out of a hat and will get to keep whichever bill he draws. The hat contains fifty $\$1$ bills, ten $\$5$ bills, and five $\$20$ bills. Find the expected value, variance, and standard deviation.

Starting with expected value:

$$\begin{array}{rl}E(x)& =\frac{50}{65}\cdot 1+\frac{10}{65}\cdot 5+\frac{5}{65}\cdot 20\\ & =3.076923\end{array}$$

To find variance, first calculate $E(X^2)$:

$$\begin{array}{rl}E(X^2)& =\frac{50}{65}\cdot 1^2+\frac{10}{65}\cdot 5^2+\frac{5}{65}\cdot 20^2\\ & =\frac{50}{65}+\frac{250}{65}+\frac{2000}{65}\\ & =\frac{2300}{65}\\ & =35.384615\end{array}$$

$$\begin{array}{rl}\text{Var}(X)& =E(X^2)-[E(X){]}^2\\ & =35.384615-(3.076923{)}^2\\ & =25.917159\end{array}$$

$$\begin{array}{rl}\text{Standard deviation}& =\sqrt{\text{Var}(X)}\\ & =\sqrt{25.917159}\end{array}$$

Solution:

Properties of expected value and standard deviation

For any two random variables $X$ and $Y$, the following properties hold. These rules make it possible to work with combined random variables without calculating everything from scratch.

Practice problem

Example:

The customer service department of a service provider receives complaints about the company’s two products: phone and internet. Here are the possible numbers of complaints of each type that the company may receive per day and their probabilities:

# of phone complaints	Probability
$0$	$0.3$
$1$	$0.5$
$2$	$0.2$

# of internet complaints	Probability
$0$	$0.4$
$1$	$0.4$
$2$	$0.2$

Part a.

Find the expected total number of complaints per day.

Let $X$ represent phone complaints and $Y$ represent internet complaints.

$$\begin{array}{rl}E(X)& =0\cdot 0.3+1\cdot 0.5+2\cdot 0.2\\ & =0.9\end{array}$$

$$\begin{array}{rl}E(Y)& =0\cdot 0.4+1\cdot 0.4+2\cdot 0.2\\ & =0.8\end{array}$$

$$\begin{array}{rl}E(X+Y)& =E(X)+E(Y)\\ & =0.9+0.8\\ & =1.7\end{array}$$

Solution:

Part b.

Assuming independence, what is the standard deviation of the total complaints?

Find the variance of each variable separately, then combine.

$$\begin{array}{rl}E(X^2)& =0^2\cdot 0.3+1^2\cdot 0.5+2^2\cdot 0.2\\ & =1.3\end{array}$$

$$\begin{array}{rl}\text{Var}(X)& =1.3-(0.9{)}^2=0.49\end{array}$$

$$\begin{array}{rl}E(Y^2)& =0^2\cdot 0.4+1^2\cdot 0.4+2^2\cdot 0.2\\ & =1.2\end{array}$$

$$\begin{array}{rl}\text{Var}(Y)& =1.2-(0.8{)}^2=0.56\end{array}$$

$$\begin{array}{rl}\text{Var}(X+Y)& =0.49+0.56=1.05\end{array}$$

$$\begin{array}{rl}\text{Standard deviation}& =\sqrt{1.05}\end{array}$$

Solution: